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snjvchan
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7^6 - 5^6 is divisible by? Updated on: 27 Jan 2019, 13:10
Originally posted by snjvchan on 27 Jan 2019, 08:08.
Last edited by Gladiator59 on 27 Jan 2019, 13:10, edited 1 time in total.
Added option E.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

75% (01:34) correct 25% (01:52) wrong based on 12 sessions

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\(7^6\) - \(5^6\) is divisible by?

(A) 6
(B) 7
(C) 16
(D) 32
(E) Both (A) and (D)

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A
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7^6 - 5^6 is divisible by? 27 Jan 2019, 08:39
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snjvchan
\(7^6\) - \(5^6\) is divisible by?

(A) 6
(B) 16
(C) 32
(D) Both (A) and (C)
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( 7^3 - 5^3 ) * (7^3 + 5^3)

343 - 125 * 343 + 125

218 * 468

Above is only / by 6

A
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7^6 - 5^6 is divisible by? 27 Jan 2019, 09:10
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snjvchan
\(7^6\) - \(5^6\) is divisible by?

(A) 6
(B) 16
(C) 32
(D) Both (A) and (C)
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(\(7^3\)-\(5^3\))[*](\(7^3\)+\(5^3\))

(343-125)*(343+125)
(218*468)


Its easy to follow now...
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7^6 - 5^6 is divisible by? 27 Jan 2019, 10:24
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solved using number properties.
7^6 - 5^6
is of form \(a^n-b^n.\) this is divisible by a+b only if n is even.
hence this will be divisible by 12. so by 6.
\(7^6 - 5^6= (7^3+5^3)(7^3-5^3)\)
\(a^n-b^n\) is divisible by n for all integers.
So the divisible should have 3. No other option has 3 as factor.
Hence only A
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7^6 - 5^6 is divisible by? 27 Jan 2019, 12:14
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snjvchan
\(7^6\) - \(5^6\) is divisible by?

(A) 6
(B) 16
(C) 32
(D) Both (A) and (C)
Show more


Hey Gladiator59 :)

how are you :) havent seen you around for quite sometime :) :lol:

In the problem above, where is option E? or GMAT format has changed :)


Another question: what algebraic formula do we use here ?

thank you and good night!!! :)
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7^6 - 5^6 is divisible by? 27 Jan 2019, 13:06
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dave13, I'm doing good. Hope you are well yourself. :-)

Looks like the OP forgot to post the option (E).

One very tricky way to solve this question would be - straightaway eliminate option (D) as if it is divisible by 32 then it must be divisible by 16 :-D

Now time to play with the exponent -

7^6 can we written as (7^3)^2 ... And so it boils down to known identity a^2-b^2 = (a+b)*(a-b)

7^6 -5^6
(7^3-5^3) * (7^3+5^3)

You could break your head with trying to remember the identity for the above but better would be to just calculate here as seven cube is 343 and five cube is 125 ....did I ever mention how useful it could be to remember cubes till 10? :-)

From this is becomes clear that 16 and hence 32 do not divide the expression and 6 does.

Good night. :-)

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